std::literals::chrono_literals::operator""ms
From cppreference.com
Defined in header <chrono>
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constexpr std::chrono::milliseconds operator ""ms( unsigned long long ms ); |
(1) | (since C++14) |
constexpr std::chrono::duration</*unspecified*/, std::milli> operator ""ms( long double ms ); |
(2) | (since C++14) |
Forms a std::chrono::duration literal representing milliseconds.
1) integer literal, returns exactly std::chrono::milliseconds(ms)
2) floating-point literal, returns a floating-point duration equivalent to std::chrono::milliseconds
Parameters
ms | - | the number of milliseconds |
Return value
The std::chrono::duration literal.
Possible implementation
constexpr std::chrono::milliseconds operator ""ms(unsigned long long ms) { return std::chrono::milliseconds(ms); } constexpr std::chrono::duration<long double, std::milli> operator ""ms(long double ms) { return std::chrono::duration<long double, std::milli>(ms); } |
Notes
This operator is declared in the namespace std::literals::chrono_literals, where both literals and chrono_literals are inline namespaces. Access to this operator can be gained with:
- using namespace std::literals,
- using namespace std::chrono_literals, or
- using namespace std::literals::chrono_literals.
In addition, within the namespace std::chrono, the directive using namespace literals::chrono_literals; is provided by the standard library, so that if a programmer uses using namespace std::chrono; to gain access to the classes in the chrono library, the corresponding literal operators become visible as well.
Example
Run this code
#include <iostream> #include <chrono> int main() { using namespace std::chrono_literals; auto d1 = 250ms; std::chrono::milliseconds d2 = 1s; std::cout << "250ms = " << d1.count() << " milliseconds\n" << "1s = " << d2.count() << " milliseconds\n"; }
Output:
250ms = 250 milliseconds 1s = 1000 milliseconds
See also
constructs new duration (public member function of std::chrono::duration<Rep,Period> ) |